Question

A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected acrose the given cell, has values of
(i) infinity (ii) $9.5\Omega$
The 'balancing lengths, on the potentiometer wire are found to be 3m and 2.85 m, respectively.The value of internal resistance of the cell is

• A. $0.25\Omega$
• B. $0.95\Omega$
• C. $0.5\Omega$
• D. $0.75\Omega$

Answer 1

Answer: (C)

$0.5\Omega$

Internal resistance of the unknown cell is
$r(\frac{l_1}{l_2}-1)R=(\frac{3}{2.85}-1)\left(9.5\Omega \right)=0.5\Omega$